TITLE: Solving $\lim_{n\to\infty}(n\int_0^{\pi/4}(\tan x)^ndx)$?
QUESTION [9 upvotes]: $$f(x)=\lim_{n\to\infty}\biggl(n\int_0^{\pi/4}(\tan x)^n\,dx\biggr)$$
I try to this way, $\tan x\ge x$, when $x\in(0,\frac\pi4)$, but this turns out to be $\tan x\ge0$, which is obvious even without calculation.
I think it can be solved by using the squeeze rule but I can't find a proper scaler to meet $A=g(x)\le f(x)\le h(x)=A$ when $n\to\infty$ 
$\color{red}{Edit}$:
According to@Khallil's advice,
I can solve that $f(n)+f(n+2)=\frac1{n+1}$, and it's trivial that $\frac1{n+1}=f(n)+f(n+2)\le 2f(n)\le f(n-2)+f(n)=\frac1{n-1}$.
So $f(x)=\frac12$.
This is easy and quick. Btw, is there any other idea?
$\color{red}{Edit[2]}$:
I find that @Byron Schmuland 's method is especially useful in a specific kind of problem. $$f(x)=\lim_{n\to\infty}\int_0^1\frac{x^n}{1+x}dx$$,for example, Let $X_,...,X_n$ be i.i.d. random variables with density$f(x)=1$ on $(0,1)$, CDF(cumulative density function) of $X$ is $F(x)=x$. Now let $M=\max(X_1,\dots,X_n)$, its density function is:
$f_M(x)=nx^{n-1}$ for $x\in(0,1)$
Also, it's also not hard to see that $M\to1$ in distribution as $n\to\infty$.
So,$$f(x)=\lim_{n\to\infty}\int_0^1\frac{x^n}{1+x}dx=\lim_{n\to\infty}\frac1n\int_0^1{f_M(x)\frac{x}{x+1}dx}=\lim_{n\to\infty}\frac1nE(\frac{M}{M+1})=0$$
Although this simple example can be solved by other ways more easily, but this gives us another perspective.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Laplace Method:

\begin{align}
&\bbox[10px,#ffd]{\lim_{n \to \infty}\bracks{n\int_{0}^{\pi/4}\tan^{n}\pars{x}\,\dd x}} =
\lim_{n \to \infty}\bracks{n\int_{0}^{\pi/4}
\tan^{n}\pars{{\pi \over 4} - x}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{n\int_{0}^{\pi/4}
\exp\pars{n\ln\pars{\tan\pars{{\pi \over 4} - x}}}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{n\int_{0}^{\infty}
\exp\pars{-2nx}\,\dd x} = \lim_{n \to \infty}\pars{n\,{1 \over 2n}} = \bbx{\ds{1 \over 2}}
\end{align}